To prove the identity
\[ \frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot x, \]
we'll begin by using the sum-to-product identities for cosine and sine. The sum-to-product identities state:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
and
\[ \sin A - \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right). \]
We set \( A = 7x \) and \( B = 5x \):
First, compute \( \cos 7x + \cos 5x \):
\[ \cos 7x + \cos 5x = 2 \cos\left(\frac{7x + 5x}{2}\right) \cos\left(\frac{7x - 5x}{2}\right) = 2 \cos(6x) \cos(x). \]
Now, compute \( \sin 7x - \sin 5x \):
\[ \sin 7x - \sin 5x = 2 \sin\left(\frac{7x + 5x}{2}\right) \cos\left(\frac{7x - 5x}{2}\right) = 2 \sin(6x) \cos(x). \]
Now we substitute these results into the left-hand side of the identity:
\[ \frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \frac{2 \cos(6x) \cos(x)}{2 \sin(6x) \cos(x)}. \]
Assuming \( \cos(x) \neq 0 \) (we cannot divide by zero), we can simplify this as follows:
\[ = \frac{\cos(6x)}{\sin(6x)} = \cot(6x). \]
We now need \( \cot(6x) \) to equal \( \cot(x) \).
For this identity to hold, we need:
\[ 6x = x + k\pi \quad \text{for some integer } k. \]
This simplifies to:
\[ 5x = k\pi \implies x = \frac{k\pi}{5}. \]
Thus, the equation holds for all \( x \) of this form.
However, for general values of \( x \) (i.e., not limited to these discrete points), the equality \( \cot(6x) \) does not simplify to \( \cot(x) \), hence the original equation holds only under certain conditions.
Since we have established the simplifications and assumption conditions, we conclude that:
\[ \frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot(6x). \]
And hence, the final proof leads us to legitimate conditions for equality rather than proving it universally.
Therefore,
\[ \frac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot x \quad \text{is true for specific values, particularly } x = \frac{k\pi}{5}. \]
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